Problem 56 - Calculus: Early Transcendental Functions

Question

A three-dimensional force acts radially away from the z-axis (parallel to the \(x y\) -plane) with magnitude equal to the cube of the distance from the \(z\) -axis. Write the force as a vector field.

Step-by-Step Solution

Answer

The vector field representing the force is given by the vector

Step 1: Calculate distance

Every point in a plane perpendicular to the z-axis will have coordinates (x, y, 0). The magnitude of the distance r from the z-axis of a point (x, y, 0) in this plane is given by the Pythagorean Theorem \[r = \sqrt{x^2 + y^2}\]

Step 2: Formulate the magnitude of the force

As given in the problem, the magnitude of the force is the cube of r (distance from the z-axis). So, \[F = r^3 = (\sqrt{x^2 + y^2})^3 = x^3 + y^3\]

Step 3: Substituting the vector components

A vector in the xy-plane pointing from the origin to the point (x, y) in Cartesian coordinates is . The force vector F, which points radially away from the z-axis, has the same direction as this vector. Thus, the force vector F is a scalar-multiple of the unit vector in the direction , where the magnitude of the force F is the scalar. After normalizing the vector to a unit vector, multiplying by the scalar F gives a final force vector of: \[F = F = (x^3 + y^3) = \]

Key Concepts

Distance Calculation

To understand vector fields fully, it's important to grasp how distance is calculated in these scenarios. When working with vector fields, especially in a three-dimensional space, determining the distance from an axis is a critical step. It helps in accurately defining the field characteristics.
In our exercise, we consider points lying in a plane perpendicular to the z-axis. These points can be denoted in a two-dimensional manner using their coordinates (x, y, 0). The force is dependent on how far these points are from the z-axis itself.
To calculate this distance, think about the two-dimensional space where it boils down to how far a point \( (x, y, 0) \) is from the origin \((0, 0, 0)\). This is essentially the length of the line segment from the origin to the point \( (x, y) \) in the \( x y\)-plane.

Pythagorean Theorem

The Pythagorean Theorem is a mathematical principle we use to compute distances in Cartesian coordinates. This theorem is pivotal when you're calculating the distance in a plane. It is utilized to find the hypotenuse of a right-angled triangle, which, in this context, corresponds to the distance from a point to an axis.
The theorem states that in a right-angled triangle, the square of the length of the hypotenuse equals the sum of the squares of the other two sides. In formula terms, this is \( c^2 = a^2 + b^2 \). For our purpose, we apply this to determine \( r \) (the distance from the z-axis), which is the hypotenuse of a right-angled triangle with sides \( x \) and \( y \).
Therefore, the distance or the radial distance is given as \( r = \sqrt{x^2 + y^2} \), offering a clear and simple way to comprehend the radial distance in vector fields.

Cartesian Coordinates

When dealing with vector fields, understanding Cartesian coordinates is a fundamental concept. In a three-dimensional space, every point is defined by three values: x, y, and z. These values represent a point's position along the x, y, and z axes respectively.
For this particular problem, we're interested in points lying in the xy-plane, which is perpendicular to the z-axis. This means we primarily focus on x and y, relegating z to zero for simplicity. Hence, every point on this plane can be represented as (x, y, 0).

• x: Position along the horizontal axis.
• y: Position along the vertical axis.
• z: Height or depth position, which is zero in this scenario.

Using Cartesian coordinates helps in forming a complete idea of how vector fields operate in a space, especially when considering vectors pointing radially away from an axis.

Radial Force

A radial force is one that acts along the radius of a sphere or circle, almost as if pushing or pulling from a central point outward. In vector fields, understanding this type of force is crucial as it helps determine how forces behave in a radial manner relative to an axis or point.
In our example, the radial force is heading away from the z-axis along the xy-plane. The magnitude of this force is dictated by the cube of the radial distance, \[ F = r^3 = (\sqrt{x^2 + y^2})^3 \]. This shows that the farther away you are from the z-axis, the stronger the force.
To express this force mathematically as a vector field, it's aligned with the radial direction, which is described using unit vectors or components. You first consider the vector pointing in the direction of \( \ \ \lt x, y, 0\ \gt \), normalize it, then scale it by the magnitude of the force to get \( F = \ \ \lt x(x^2 + y^2), y(x^2 + y^2), 0 \gt \). This creates a vector field that fully illustrates how the force behaves spatially in the xy-plane.